Given: ${\overrightarrow{u}}_1=\hat{i}+\hat{j}+a\hat{k}, {\overrightarrow{u}}_2=\hat{i}+b\hat{j}+\hat{k}$ and ${\overrightarrow{u}}_3=c\hat{i}+\hat{j}+\hat{k}$ are coplanar.

Also given that,${\overrightarrow{v}}_1=\left(a+b\right)\hat{i}+c\hat{j}+c\hat{k}, {\overrightarrow{v}}_2=a\hat{i}+\left(b+c\right)\hat{j}+a\hat{k}$and ${\overrightarrow{v}}_3=b\hat{i}+b\hat{j}+\left(c+a\right)\hat{k}$ are coplanar.

Now, using the condition of coplanar we get,

$\Rightarrow \left|\begin{array}{ccc}1& 1& a\\ 1& b& 1\\ c& 1& 1\end{array}\right|=0$

Expanding the determinant along $R_1$.

$\Rightarrow \left(b-1\right)-\left(1-c\right)+a\left(1-bc\right)=0$

$\Rightarrow a+b+c=2+abc .....\left(\mathrm{i}\right)$

Again using the coplanar condition we get,

$\Rightarrow \left|\begin{array}{ccc}a+b& c& c\\ a& b+c& a\\ b& b& c+a\end{array}\right|=0$

Apply row transformations, $\left(R_3\to R_3-\left(R_1+R_2\right)\right)$

$\Rightarrow \left|\begin{array}{ccc}a+b& c& c\\ a& b+c& a\\ -2a& -2c& 0\end{array}\right|=0$

Expand the determinant along $R_1$.

$\Rightarrow \left(a+b\right)\left(0+2ac\right)-c\left(0+2a^2\right)+c\left(-2ac+2a\left(b+c\right)\right)=0$

$\Rightarrow 2a^{2}c+2abc-2a^{2}c-2a{c}^2+2abc+2a{c}^2=0$

$\Rightarrow abc=0$

$\therefore a+b+c=2$ (From eq $\left(i\right)$)

$\therefore 6\left(a+b+c\right)=12$

Hence this is the correct option.